Consider the RLC circuit shown below The general form of the ODE for this type o

Question

Consider the RLC circuit shown below The general form of the ODE for this type of circuit, describing the charge on the capacitor, V_e is LC d^2V_c/dt^2 + RC dV_c/dt + V_c = V_in a)(*) Substitute the given values, and determine the general homogeneous solution to the ODE by finding and solving the characteristic equation. For each of the following situations, assume that the capacitor begins uncharged (V_c(0) = 0) and the initial current is zero (I(0) = V'_c(0) = 0), and use the listed methed to solve the ODE for V_c: b)(*) The input voltage is given by V_in = 2e^-2t cost Solve the ODE using the Methed of Undetermined Coefficients. c)(**) The input voltage is given by Vin = 2e=2e^-2t cos t Solve the ODE using the Methed of Undetermined Coefficients. d)(***)The input voltage is given by V_in = e^-4t Solve the ODE using the Methed of Undetermined Coefficients. e)(**) The input voltage is given by V_in ={10 t

Explanation / Answer

part a:

using values of R,L and C in the homogenous part of the ODE:

0.05*Vc'' + 0.45*Vc' + Vc =0

where Vc'=dVc/dt

and Vc''= d^2 Vc/dt^2

let Vc=e^(m*t)

then characteristics equation becomes:

0.05*m^2+0.45*m+1=0

==>m=-4 or m=-5

then solution for homogenous equation:

Vc(t)=A*exp(-4*t)+B*exp(-5*t)

where A and B are two constants.

part b:

method of undetermined coefficients:

given that in(t)=6*e^(-t)

then let Vc(t)=C*exp(-t)

then Vc'(t)=-C*exp(-t)

Vc''(t)=C*exp(-t)

using the first and second derivatives in complete ODE:

0.05*C(exp(-t)-0.45*C*exp(-t)+C*exp(-t)=6*exp(-t)

==>0.6*C=6

==>C=6/0.6=10

hence complete expression for Vc(t)=A*exp(-4*t)+B*exp(-5*t)+10*exp(-t)

given that Vc(t)=0

==>A+B+10=0

==>A+B=-10....(1)

given that Vc'(t)=0

==>-4*A-5*B-10*1=0

==>4*A+5*B=-10...(2)

solving 1 and 2 we get

A=-40

B=30

hence final solution for Vc(t)=-40*exp(-4*t)+30*exp(-5*t)+10*exp(-t)

part c:

solution to the homogenous part will remain the same as Vc(t)=A*exp(-4*t)+B*exp(-5*t)

let particular solution be Vc(t)=exp(-2*t)*(C*cos(t)+D*sin(t))

Vc'(t)=exp(-2t)*(-C*sin(t)+D*cos(t))-2*exp(-2*t)*(C*cos(t)+D*sin(t)))

=exp(-2*t)*(-(C+2*D)*sin(t)+(D-2*C)*cos(t))

Vc''(t)=-exp(-2*t)*((C+2*D)*cos(t)+(D-2*C)*sin(t)) -2*exp(-2*t)*(-(C+2*D)*sin(t)+(D-2*C)*cos(t))

=-exp(-2*t)*((3*D-C)*cos(t)-(3*C+D)*sin(t))

using the expression in the original ODE:

0.05*Vc''+0.45*Vc'+Vc=Vin

==>0.05*(-exp(-2*t)*((3*D-C)*cos(t)-(3*C+D)*sin(t)))+0.45*(exp(-2*t)*(-(C+2*D)*sin(t)+(D-2*C)*cos(t)))

+(exp(-2*t)*(C*cos(t)+D*sin(t))) =2*exp(-2*t)*cos(t)

comparing the coefficient of cos(t) and sin(t):

-0.05*(3*D-C)+0.45*(D-2*C)+C=2

==>0.15*C+0.3*D=2...(3)

and 0.05*(3*C+D)-0.45*(C+2*D)+D=0

==>-0.3*C+0.15*D=0...(4)

solving equation 3 and equation 4 , we get

C=2.67

D=5.333

hence complete solution:

Vc(t)=A*exp(-4*t)+B*exp(-5*t)+exp(-2*t)*(2.67*cos(t)+5.333*sin(t))

at t=0, vc(t)=0

==>A+B+2.67=0

==>A+B=-2.67...(5)

at t=0, Vc'(t)=0

==>-4*A-5*B=0..(6)

solving equation 5 and 6, we get

A=-13.35

B=10.68

hence final solution: Vc(t)=-13.35*exp(-4*t)+10.68*exp(-5*t)+exp(-2*t)*(2.67*cos(t)+5.333*sin(t))

part D:

let particular solution be Vc(t)=C*t*exp(-4*t)

Vc'=C*exp(-4*t)-4*C*t*exp(-4*t)

=C*exp(-4*t)*(1-4*t)

Vc''=C*exp(-4*t)*(-4)-4*C*exp(-4*t)*(1-4*t)

=C*exp(-4*t)*(-4-4+16*t)

=C*exp(-4*t)*(16*t-8)

substituing in original ODE:

0.05*(C*exp(-4*t)*(16*t-8))+0.45*(C*exp(-4*t)*(1-4*t))+C*t*exp(-4*t)=exp(-4*t)

==>0.05*C*(16*t-8)+0.45*C*(1-4*t)+C*t=1

==>0.05*C=1

==>C=1/0.05=20

hence complete expression: Vc(t)=A*exp(-4*t)+B*exp(-5*t)+20*t*exp(-4*t)

=(A+20*t)*exp(-4*t)+B*exp(-5*t)

Vc'(t)=(A+20*t)*(-4)*exp(-4*t)+20*exp(-4*t)-5*B*exp(-5*t)

given that Vc(0)=0

==>A+B=0..(7)

Vc'(0)=0

==>-4*A+20-5*B=0

==>4*A+5*B=20..(8)

solving equation 7 and 8:

A=-20

B=20

hence complete solution:

Vc(t)=(-20+20*t)*exp(-4*t)+20*exp(-5*t)

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